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(F)=5/3.F=
We move all terms to the left:
(F)-(5/3.F)=0
Domain of the equation: 3.F)!=0We add all the numbers together, and all the variables
F!=0/1
F!=0
F∈R
F-(+5/3.F)=0
We get rid of parentheses
F-5/3.F=0
We multiply all the terms by the denominator
F*3.F-5=0
Wy multiply elements
3F^2-5=0
a = 3; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·3·(-5)
Δ = 60
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*3}=\frac{0-2\sqrt{15}}{6} =-\frac{2\sqrt{15}}{6} =-\frac{\sqrt{15}}{3} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*3}=\frac{0+2\sqrt{15}}{6} =\frac{2\sqrt{15}}{6} =\frac{\sqrt{15}}{3} $
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